Problem 68 - Calculus - AP Edition

Question

(a) Explain why L'Hôpital's rule does not apply to the $$\lim _{x \rightarrow 0} \frac{x^{2} \sin (1 / x)}{\sin x}$$ (b) Find the limit.

Step-by-Step Solution

Answer

L'Hôpital's Rule is inapplicable because \( x^2 \sin(1/x) \) is not differentiable at \( x=0 \). The limit is 0 by the Squeeze Theorem.

Step 1: Evaluate Direct Substitution

Evaluate the limit \( \lim _{x \rightarrow 0} \frac{x^{2} \sin (1 / x)}{\sin x} \) by directly substituting \( x = 0 \). The expression becomes \( \frac{0}{0} \), which is an indeterminate form and suggests that direct substitution doesn't work. This often indicates a potential for using L'Hôpital's Rule, but other checks are needed.

Step 2: Check the Conditions for L'Hôpital's Rule

L'Hôpital's Rule applies if the limit is of the form \( \frac{0}{0} \) or \( \frac{\pm\infty}{\pm\infty} \) after substituting the appropriate values and the functions in the numerator and denominator are differentiable. Here:- The numerator \( x^2 \sin(1/x) \) around \( x=0 \) is not differentiable at \( x=0 \) due to \( \sin(1/x) \) being highly oscillatory and undefined directly in terms of a derivative at \( x = 0 \). - Therefore, L'Hôpital's Rule does not apply.

Step 3: Simplify the Expression for Limit Calculation

Notice that \( |\sin(1/x)| \leq 1 \). Therefore, \( x^2 |\sin(1/x)| \leq x^2 \), leading to:\[ \left| \frac{x^2 \sin(1/x)}{\sin x} \right| = \frac{x^2|\sin(1/x)|}{|\sin x|} \leq \frac{x^2}{|\sin x|} \]As \( x \to 0 \), \( \sin x \approx x \). Thus, \( \frac{x^2}{\sin x} \to x \).

Step 4: Apply Squeeze Theorem

Since \( \left| \frac{x^2 \sin(1/x)}{\sin x} \right| \leq |x| \) and as \( x \to 0 \), \( |x| \to 0 \), by the Squeeze Theorem, \(\lim _{x \rightarrow 0} \frac{x^2 \sin(1/x)}{\sin x} = 0. \)

Key Concepts

Indeterminate Forms

An indeterminate form appears in calculus when the limits of certain mathematical expressions are not immediately clear. These forms suggest ambiguity in calculations and require us to use special rules or theorems to accurately determine a limit. Common indeterminate forms include:

• \( \frac{0}{0} \)
• \( \frac{\infty}{\infty} \)
• \( 0 \cdot \infty \)
• \( \infty - \infty \)

It's crucial to identify these forms because they are prevalent in calculus problems involving limits. In our example, when trying to solve \( \lim _{x \rightarrow 0} \frac{x^{2} \sin (1 / x)}{\sin x} \), direct substitution leads to\( \frac{0}{0} \). This hints at an indeterminate form. However, even though L'Hôpital's Rule can often solve such cases, it requires differentiability of both the numerator and denominator, which will not occur here.

Squeeze Theorem

The Squeeze Theorem is a powerful tool in limit evaluation. It allows us to deduce the limit of a function if that function is "squeezed" between two others that have a known limit. Here's how it works:

• If \( f(x) \leq g(x) \leq h(x) \) for all \( x \) near \( a \) (except possibly at \( a \) itself).
• And \( \lim_{x \to a} f(x) = \lim_{x \to a} h(x) = L \).
• Then \( \lim_{x \to a} g(x) = L \) too.

In our solution, we observe \( |\sin(1/x)| \leq 1 \), hence \( x^2 |\sin(1/x)| \leq x^2 \). So, \[ \left| \frac{x^2 \sin(1/x)}{\sin x} \right| \leq \frac{x^2}{|\sin x|} \]. If, as \( x \to 0 \), \( \frac{x^2}{\sin x} \approx x \to 0 \), then by the Squeeze Theorem, the original limit also approaches 0.

Differentiability

Differentiability is a fundamental concept in calculus that refers to the ability of a function to have a derivative at a certain point. A function being differentiable means it must be smooth enough at that point - essentially, no sharp turns, breaks, or vertical tangents.
For L'Hôpital's Rule, this concept is critical as the functions involved must be differentiable at the limit point. In this exercise's problem, the numerator \( x^2 \sin(1/x) \) involves \( \sin(1/x) \), which becomes highly oscillatory and undefined directly around \( x=0 \). The lack of a well-defined slope (derivative) here makes the function non-differentiable at that point. Since both functions in the fraction must satisfy this differentiability condition for L'Hôpital's Rule to apply, we see why the rule isn't applicable in this particular exercise.